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Standard X

Mathematics

Calculating Heights and Distances

The top of a ...

Question

The top of a hill observed from the top and bottom of a building of height $h$ is at angles of elevation $α$ and $β$ respectively. The hehill is:ight of the

\frac{h cot β}{cot β - cot α}

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\frac{h cot α}{cot α - cot β}

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\frac{h tan α}{tan α - tan β}

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None of the above

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Solution

The correct option is C $\frac{h cot α}{cot α - cot β}$
Refer the image
AB is the hill and CD is the building
$tan (α) = \frac{A B}{B E} ⟹ B E = A B cot (α) tan (β) = \frac{A B}{B D} ⟹ B D = A B cot (β)$
Also, $tan (α) = \frac{C D}{D E}$
$⟹ D E = C D cot (α) = h cot (α)$
Now, $B E = B D + D E$
$\Rightarrow A B cot (α) = A B cot (β) + h cot (α) \Rightarrow A B = \frac{h cot (α)}{cot (α) - cot (β)}$
Option B is correct

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The top of a hill observed from the top and bottom of a building of height h is at angles of elevation α and β respectively. The hehill is:ight of the

The correct option is C hcotαcotα−cotβRefer the imageAB is the hill and CD is the buildingtan(α)=ABBE⟹BE=ABcot(α)tan(β)=ABBD⟹BD=ABcot(β) Also, tan(α)=CDDE⟹DE=CDcot(α)=hcot(α) Now, BE=BD+DE⇒ABcot(α)=ABcot(β)+hcot(α)⇒AB=hcot(α)cot(α)−cot(β)Option B is correct

The top of a hill observed from the top and bottom of a building of height $h$ is at angles of elevation $α$ and $β$ respectively. The hehill is:ight of the